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Suppose an operator $T$ is bounded on $L^2$ and also bounded from $L^{1}$ to $L^{1}$-weak. Then by Marcinkewicz interpolation one gets that $T$ is bounded on every $L^{p}$ for p between 1 and 2. Precise versions of the theorem (see the book of L.Grafakos) give an estimate of the norm of $T$ on $L^{p}$, and of course the norm diverges as $p\to 1$. If not, by a simple argument one could obtain that $T$ is bounded also on $L^{1}$, which is clearly false in general (e.g. singular integrals). By the way, even a weaker assumption on the growth of the norm should allow one to conclude that $T$ is bounded on $L^{1}$, as in extrapolation theorems.

Now suppose you (me) are hard-headed and want to use the general machinery of real interpolation, say the K-method as detailed in the book of Bergh and Löfström. Then $T$ is bounded from $L^{1}$ to $L^{1}$-weak which means the Lorentz space $L^{1,\infty}$, with norm $M_{0}$, and also from $L^{2}$ to $L^{2}$ with norm $M_{1}$. Then $T$ is bounded on the corresponding real interpolation spaces with norm $M_{0}^{1-\theta}M_{1}^{\theta}$. Real interpolates of Lorentz spaces are Lorentz spaces, see Theorem 5.3.1 in B-L. We conclude that $T$ is bounded on every Lorentz space $L^{p,q}$ with p between 1 and 2, and $1\le q\le \infty$, and in particular on $L^{p,p}=L^{p}$ as expected.

Unfortunately, now I have a uniform bound on the norm of $T$ as $p\to 1$, which would allow me to conclude that $T$ is also bounded on $L^{1}$. Where is the mistake? There must be some inaccuracy in one of the steps, but which one exactly?

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1 Answer 1

up vote 6 down vote accepted

(Note: I don't have my copy of B&L handy, so I'm sort of doing this from memory.)

The problem is that $(L_{p_0.q_0},L_{p_1,q_1})_{\theta,q} = L_{p_\theta,q}$ under equivalent norm. There's no saying what the constant of equivalency are. Now, if you look at $T: \{X_1,X_2\} \to \{Y_1,Y_2\}$ with norm $\{M_1,M_2\}$, then you have $$ T: (X_1,X_2)_{\theta,q} \to (Y_1,Y_2)_{\theta,q} $$ with norm $M_1^{(1-\theta)}M_2^\theta$. Now put in $X_1 = L_{1,1}$, $X_2 = L_{2,2}$, $Y_1 = L_{1,\infty}$ and $Y_2 = L_{2,2}$, while the interpolants $(X_1,X_2)_{\theta,q}$ may be equivalent to $L_{p_\theta,q}$, and $(Y_1,Y_2)_{\theta,q}$ to $L_{p_\theta,q}$, the constants of equivalency may be different. That's where the degeneracy is hiding.

Observe that if $Y_1$ were $L_{1,1}$ (or if $X_1 = L_{1,\infty}$) instead, then the two interpolation spaces are actually equal, which is what you expect.

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You may also find ams.org/mathscinet-getitem?mr=169037 and ams.org/mathscinet-getitem?mr=178381 interesting and useful. –  Willie Wong Jul 4 '10 at 11:33
    
So basically this argument shows that the equivalence between $L^p$ and the real interpolation space is modulo a constant that grows to infinity as $p\to$, and can even be estimated. This settles the question I guess –  Piero D'Ancona Jul 4 '10 at 15:19

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